Môđun của số phức z thỏa mãn $\left| z-1 \right|=5$ và $17\left(...

Đặt $z=a+bi\left( a;b\in \mathbb{R} \right)$.
Ta có: $\begin{aligned}
& \left\{ \begin{aligned}
& \left| z-1 \right|=5 \\
& 17\left( z+\overline{z} \right)-5.z.\overline{z}=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{\left( a-1 \right)}^{2}}+{{b}^{2}}=25 \\
& 17.2a-5\left( {{a}^{2}}+{{b}^{2}} \right)=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& ({{a}^{2}}+{{b}^{2}})-2a-24=0 \\
& 17.2a-5\left( {{a}^{2}}+{{b}^{2}} \right)=0 \\
\end{aligned} \right. \\
& \Leftrightarrow \left\{ \begin{aligned}
& 5\left[ \left( {{a}^{2}}+{{b}^{2}} \right)-2a-24 \right]=0 \\
& 17.2a-5\left( {{a}^{2}}+{{b}^{2}} \right)=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& 34a+5\left( -2a-24 \right)=0 \\
& 5\left( {{a}^{2}}+{{b}^{2}} \right)=17.2a \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=5 \\
& {{a}^{2}}+{{b}^{2}}=34 \\
\end{aligned} \right. \\
\end{aligned}$
Suy ra: $\left| z \right|=\sqrt{{{a}^{2}}+{{b}^{2}}}=\sqrt{34}$.